This is the first in a series of posts about the Atiyah-Singer index theorem.

Let {V, W} be finite-dimensional vector spaces (over {\mathbb{C}}, say), and consider the space {\hom_{\mathbb{C}}(V, W)} of linear maps {T: V \rightarrow W}. To each {T \in \hom_{\mathbb{C}}(V, W)}, we can assign two numbers: the dimension of the kernel {\ker T} and the dimension of the cokernel {\mathrm{coker} T}. These are obviously nonconstant, and not even locally constant. However, the difference {\dim \ker T - \dim \mathrm{coker} T = \dim V - \dim W} is constant in {T}.

This was a trivial observation, but it leads to something deeper. More generally, let’s consider an operator (such as, eventually, a differential operator), on an infinite-dimensional Hilbert space. Choose separable, infinite-dimensional Hilbert spaces {V, W}; while they are abstractly isomorphic, we don’t necessarily want to choose an isomorphism between them. Consider a bounded linear operator {T: V \rightarrow W}.

Definition 1 {T} is Fredholm if {T} is “invertible up to compact operators,” i.e. there is a bounded operator {U: W \rightarrow U} such that {TU - I} and {UT - I} are compact.

In other words, if one forms the category of Hilbert spaces and bounded operators, and quotients by the ideal (in this category) of compact operators, then {T} is invertible in the quotient category. It thus follows that adding a compact operator does not change Fredholmness: in particular, {I + K} is Fredholm if {V = W} and {K: V \rightarrow V} is compact.

Fredholm operators are the appropriate setting for generalizing the small bit of linear algebra I mentioned earlier. In fact,

Proposition 2 A Fredholm operator {T: V \rightarrow W} has a finite-dimensional kernel and cokernel.

Proof: In fact, let {V' \subset V } be the kernel. Then if {v' \in V'}, we have

\displaystyle v' = UT v' + (I - UT) v' = (I - UT) v'

where {U} is a “pseudoinverse” to {T} as above. If we let {v'} range over the elements of {v'} of norm one, then the right-hand-side ranges over a compact set by assumption. But a locally compact Banach space is finite-dimensional, so {V'} is finite-dimensional. Taking adjoints, we can similarly see that the cokernel is finite-dimensional (because the adjoint is also Fredholm). \Box

The space of Fredholm operators between a pair of separable, infinite-dimensional Hilbert spaces is interesting. For instance, it has the homotopy type of {BU \times \mathbb{Z}}, so it is a representing space for K-theory. In particular, the space of its connected components is just {\mathbb{Z}}. The stratification of the space of Fredholm operators is given by the index.

Definition 3 Given a Fredholm operator {T: V \rightarrow W}, we define the index of {T} to be {\dim \ker T - \dim \mathrm{coker} T}. (more…)

Back to elliptic regularity. We have a constant-coefficient partial differential operator {P = \sum_{a: |a| \leq k} C_a D^a} which is elliptic, i.e. the polynomial

\displaystyle Q(\xi) = \sum_{a: |a| \leq k} C_a \xi^a

satisfies {|Q(\xi)| \geq \epsilon |\xi|^k} for {|\xi|} large. We used this last property to find a near-fundamental solution to {P}. That is, we chose {E} such that {\hat{E} = (1-\varphi) Q^{-1}}, where {\varphi} was our arbitrary cut-off function equal to one in some neighborhood of the origin. The point of all this was that

\displaystyle P(E) = \delta - \hat{\varphi}.

In other words, {E} is near the fundamental solution. So given that {Pf = g}, we can use {E} to “almost” obtain {f} from {g} by convolution {E \ast g}—if this were exact, we’d have the fundamental solution itself.

We now want to show that {E} isn’t all that badly behaved.

The singular locus of the parametrix

We are going to show that {\mathrm{sing} E = \{0\}}. The basic lemma we need is the following. Fix {m}. Consider a smooth function {\phi} such that, for each {a}, there is a constant {M_a} with

\displaystyle |D^a \phi(x)| \leq M_a (1+|x|)^{m-|a|};

then this is a distribution, but it is not necessarily a Schwarz function. And {\hat{\phi}} cannot be expected to be one, thus. Nevertheless:

Lemma 1 {\hat{\phi}} is regular outside the origin. (more…)